Answer to Question #254449 in Statistics and Probability for Goldie

Question #254449

The management of Resale Furniture, a change of second hand furniture stores in Metro Manila, designed an incentive plan for salespeople. To evaluate this innovative incentive plan, 10 salespeople were selected at random, and their weekly incomes before and after the incentive plan were recorded. 

SALE PERSON 1

Before: 2000

After: 3500

SALE PERSON 2

Before: 3000

After:4120

SALE PERSON 3

Before: 2500

After: 3800

SALE PERSON 4

Before: 3100

After: 4200

SALE PERSON 5

Before: 1900

After: 1820

SALE PERSON 6

Before: 1750

After: 1600

SALE PERSON 7

Before: 2300

After: 3400

SALE PERSON 8

Before: 3210

After: 1900

SALE PERSON 9

Before: 2340

After: 2340

SALE PERSON 10

Before: 1870

After: 3290

Was there a significant increase in the typical salesperson's weekly income due to the innovative incentive plan? Use the 0.05 level of significance.


1
Expert's answer
2021-10-22T00:46:52-0400

The two samples are dependent because each pair of figures apply to one person. Therefore, to perform this test, we shall use the paired t-test to make a decision as described below.

The hypotheses to be tested are,

"H_0:\\mu_d=0"

"Against"

"H_1:\\mu_d\\gt0", where "d" is the difference between each pair of observations.

To test these hypotheses, we first determine the mean of the differences"(\\bar{d})" given by,

"\\bar{d}=(\\sum(d))\/n", where "n=10."

Now,

"\\sum(d)=-6000"

Therefore,

"\\bar{d}=-6000\/10=-600"

Variance of the differences is given by,

"V(d)=\\sum(d-\\bar{d})^2\/(n-1)"

"V(d)=299375800\/9=" 33263978

The standard deviation "SD(d)" is given by,

"SD(d)=\\sqrt{V(d)}=\\sqrt{33263978}=5767.493197"

The test statistic is given by,

"t^*=\\bar{d}\/SD(d)\/\\sqrt{n}=-600\/(5767.493197\/\\sqrt{10})=-600\/1823.841489=-0.329 (2\\space dp)."

"t^*" is compared with the table value with (n-1)=10-1=9 degrees of freedom at "\\alpha=0.05" level of significance.

The table value "(t)" is,

"t_{\\alpha, 9}=t_{0.05,9}=1.833" and the null hypothesis is rejected if "|t^*|\\gt t_{0.025,9}"

Since "|t^*|=|-0.329|=0.329\\lt t_{0.025,9}=1.833", we fail to reject the null hypothesis and conclude that there is no sufficient evidence to show that there was a significant increase in typical salesperson's weekly income due to the innovative incentive plan at 5% level of significance.


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